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Chapter 3: Problem 65

Consider two projectiles launched on level ground with the same speed, at angles \(45^{\circ} \pm \alpha .\) Show that the ratio of their flight times is \(\tan \left(\alpha+45^{\circ}\right)\)

Short Answer

Expert verified
The ratio of flight times of two projectiles launched at angles \(45^{\circ} \pm \alpha \) is indeed given by \(\tan \left(\alpha+45^{\circ}\right)\).

Step by step solution

01

Find the formula for time of flight

Projectiles follow a path determined by the force of gravity pulling them back to the ground and their initial velocity. The formula for time of flight (T) of a projectile is given by \[T=\frac{2u\sin \theta}{g}\], where \(u\) is the initial velocity, \(\theta\) is the angle of launch and \(g\) is acceleration due to gravity. For the two projectiles in this problem, both are launched with the same speed \(u\), but at different angles. Therefore the times of flight will be \(T_1=\frac{2u\sin (45^{\circ}+\alpha)}{g}\) and \(T_2=\frac{2u\sin (45^{\circ}-\alpha)}{g}\).
02

Calculate the ratio of flight times

The ratio of the two flight times \(T_1\) and \(T_2\) is given by: \[ \frac{T_1}{T_2} = \frac{\frac{2u\sin (45^{\circ}+\alpha)}{g}}{\frac{2u\sin (45^{\circ}-\alpha)}{g}}. \] Simplifying this will give us: \[ \frac{\sin (45^{\circ}+\alpha)}{\sin (45^{\circ}-\alpha)}. \] We will use the trigonometric identity for \(\sin(a+b)\) and \(\sin(a-b)\) which are \(\sin(a+b)= \sin a \cos b + \cos a \sin b\) and \(\sin(a-b)= \sin a \cos b - \cos a \sin b\) respectively to simplify the ratio.
03

Apply the trigonometric identities

Applying the identities we get: \[ \frac{\sin 45^{\circ}\cos \alpha + \cos 45^{\circ}\sin \alpha}{\sin 45^{\circ}\cos \alpha - \cos 45^{\circ}\sin \alpha} = \frac{(\cos \alpha + \sin \alpha)}{(\cos \alpha - \sin \alpha)}. \] Since \(\cos 45^{\circ} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}\), we substitute these into the equation, yielding: \[ \frac{T_1}{T_2} = \tan \left(45^{\circ} + \alpha\right). \] This is what was required to be shown.

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